What's new

# EWMA

#### Bester

##### Member
Subscriber
In the following question, I agree that b is the correct answer, but why is "a" not also correct. The EWMA do not include the mean reversion term (i.e product of weight and long run variance). But does this not indirectly assume that the long run volatility/variance is zero in EWMA, therefore it is not included.

Which of the following statements about the exponentially weighted moving average (EWMA) model and the generalized autoregressive conditional heteroscedasticity (GARCH(1,1)) model is correct?

a. The EWMA model is a special case of the GARCH(1,1) model with the additional assumption that the longrun volatility is zero.
b. A variance estimate from the EWMA model is always between the prior day’s estimated variance and the
prior day’s squared return.
c. The GARCH(1,1) model always assigns less weight to the prior day’s estimated variance than the EWMA model.
d. A variance estimate from the GARCH(1,1) model is always between the prior day’s estimated variance and
the prior day’s squared return.

Correct answer: b

Explanation: The EWMA estimate of variance is a weighted average of the prior day’s variance and prior day
squared return.

#### ami44

##### Well-Known Member
Subscriber
I agree, that looks like an error in the question to me.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
But (a) is incorrect. The additional term in GARCH is the product of the long-run variance and its weight (in Hull, omega is the term which is the product and gamma is the weight; gamma = 1 - alpha - beta). If we take a GARCH(1,1) model and add "the additional assumption that the longrun volatility is zero" we do not end up with an EWMA because alpha + beta < 1.0; i.e., there is still a "pull" toward zero which the EWMA does not have. A more true statement would be something like "the EWMA model is a special case of the GARCH(1,1) model with the additional assumption that the weight given to the longrun volatility is zero." Thanks,

#### Bester

##### Member
Subscriber
Thanks David for the quick response

#### ami44

##### Well-Known Member
Subscriber
Wow, that is a subtle difference, but you are right of course.

#### Matthew Graves

##### Active Member
Subscriber
I had to do a double take on this question as well. As David says, the subtle wording is key. EWMA doesn't assume anything about the long run variance since it doesn't take it into account at all!

#### akrushn2

##### Member
I had a as my answer and I can see why thats not correct. but can someone please clarify on choice b and d further? I dont understand the wording as between previous day's variance and previous day's squared return? the variance estimate is the sum of both as per the ewma variance formulas. what am i missing here?

Last edited:

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @akrushn2 Here is an example using exaggerated numbers. Assume EMWA's lambda, λ = 0.80 (which is just a weight) and the prior variance estimate (because volatility was 10.0%) is 10%^2 = 0.010
• If the return is higher at 15.0%, then squared return is 15.0%^2 and the updated EWMA variance estimate is given by σ^2 = 0.80*10%^2 + 0.20*15%^2 = 11.18%^2; the weighted 80/20 average of 10%^2 and 15%^2 must fall somewhere between them!
• If the return is higher at 5.0%, then squared return is 5.0%^2 and the updated EWMA variance estimate is given by σ^2 = 0.80*10%^2 + 0.20*5%^2 = 9.22%%^2; the weighted 80/20 average of 10%^2 and 5%^2 must fall somewhere between them!
So abstractly, this is really just the idea that X*A + (1-X)*B must lie somewhere between A and B because it's a weighted average of A and B; e.g., X*3 + (1-X)*7 must lie between 3 and 7.

This would be also true of GARCH(1,1) if we were only weighting by alpha and beta, but GARCH(1,1) is a weighted average additionally of the long-run variance which can be in-between the other two values or could be outside them (above or below) so statement (D) won't be necessarily true for GARCH(1,1); even as it will often be true, because the gamma weight is so small. Thanks,

#### akrushn2

##### Member
Well understood. Thanks David